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Mariulka [41]
3 years ago
15

If the mass of a material is 41 grams and the volume of the material is 5 cm^3, what would the density of the material be? answe

r: Units:
Physics
2 answers:
Mila [183]3 years ago
6 0
Density can be calculated using the following rule:
Density = mass/volume

We are given that:
mass = 41 grams
volume = 5 cm^3

Substitute with the givens in the above equation to get the density as follows:
Density = 41/5 = 8.2 g/cm^3

Based on the above calculations:
answer : 8.2
units : g/cm^3
sergejj [24]3 years ago
6 0
Density= m/v   ... thus substitute in your values and you get density= (41)/(5)... which basically is equally to 8.2
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GREYUIT [131]

Answer:

?????...............

4 0
3 years ago
Read 2 more answers
Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 pis
Vikki [24]

Answer:

1) Final Temperature of the gas in A will be GREATER than the temperature in B

2) Diagram of both processes on a single PV has been uploaded below

3) The Initial  pressures in containers A and B is 3039.87 J/liters

4) the final volume of container B is 923.36 cm³

Explanation:

Given that;

Temperature = 20°C = 293 K

mass of piston = 10 kg

Area = 100cm³

Volume V = 800 cm³ = 0.8 L

ideal gas constant R = 8.3 J/K·mol

1)

Final Temperature of the gas in A will b GREATER than the temperature in B

2)

Diagram of both processes on a single PV has been uploaded below,

3)

Initial  pressures in containers A and B

PV = nRT

P = RT/V

we substitute

P = (8.3 × 293) /  0.8

P = 2431.9 / 0.8

P = 3039.87 J/liters

Therefore, The Initial  pressures in containers A and B is 3039.87 J/liters

4)

Given that;

power = 25 W

time t = 15s

the final volume of container B = ?

we know that;

work done = power × time

work done = 25 × 15 = 375

Also work done = P( V₂ - V₁ )

so we substitute

375 = 3039.87 ( V₂ - 0.8 )

( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³

7 0
3 years ago
A __ allows you to determine the position of an object <br><br>this is science plz helpp​
Alex_Xolod [135]
The correct answer is C
8 0
3 years ago
Must be 8 letters (Dimensional Analysis escape room)
makkiz [27]

Answer: A-B-C-E-G-K-H-I

Explanation:

Ok, we start at A.

16 pints = x gallons.

We know that:

1 pint = 0.125 al

then:

16pint = 16*(0.125gal) = 2 gal.

The second letter is B.

Now we have:

12,320 yd = X mi

And we know that:

1 mi = 1760 yd.

Then 12,320yd = (12,320/1760) mi = 7mi

the letter is C.

Now we have:

432 cups = X gals

and:

1 cup = 0.0625 gal

then 432 cups = 432*(0.0625 gal) = 27 gal.

The next letter is E.

Now we have:

1/4 ton = X pounds.

And 1 ton = 2000 pounds

then 1/4 ton = (1/4)*2000 pounds = 500 pounds

The next letter is G.

Now we have:

3.5 gal = X ounces.

and we know that:

1 gal = 128 oz

Then:

3.5 gal = 3.5*(128oz) = 448oz

The next letter is K.

Now we have:

1/2 day = x minutes.

and we know that:

1 day = 24*60 min = 1440 min

then 1/2 day = (1/2)*1440 min = 720 min.

The next letter is H.

Now we have:

360 in = x yd

And we know that:

1yd = 36 inches.

then 360 in = (360/36) = 10 yd.

The next letter is I

Now we have:

56 cups = x quarts.

and we know that;

1cup = 0.25 quart

then: 56 cups = 56*0.25 quart = 14

So here this ends.

Then the code is:

A-B-C-E-G-K-H-I

7 0
3 years ago
Two spherical asteroids have the same radius R. Asteroid 1 hasmass M and asteroid 2 has mas 2M. The two asteroids are releasedfr
gtnhenbr [62]

Answer:

 v_1 =\sqrt{\dfrac{16GM}{15R}}

 v_2 =\sqrt{\dfrac{4GM}{15R}}

Explanation:

given,

mass of asteroid 1 = M

mass of asteroid 2 = 2M

radius of two asteroid = R

Distance between the asteroid = 10 R

Speed of the asteroid before collision = ?

using conservation of momentum

M u + 2M u' = M v₁ + 2 M v₂

initial speed of asteroid is equal to zero

0 = v₁ + 2 v₂

v₁ = -2 v₂

using conservation of momentum

initial potential energy is converted into potential energy and the kinetic energy of both the asteroids.

 \dfrac{GM(2M)}{10R}=\dfrac{GM(2M)}{2R}+\dfrac{1}{2}Mv_1^2 + \dfrac{1}{2}(2M)v_2^2

 \dfrac{GM(2M)}{10R}-\dfrac{GM(2M)}{2R}=\dfrac{1}{2}M(-2v_2)^2 + \dfrac{1}{2}(2M)v_2^2

 6v_2^2 = \dfrac{8GM}{5R}

 v_2 =\sqrt{\dfrac{4GM}{15R}}

now,

 v_1 =-2\sqrt{\dfrac{4GM}{15R}}

 v_1 =\sqrt{\dfrac{16GM}{15R}}

hence, the velocity of asteroid are

 v_1 =\sqrt{\dfrac{16GM}{15R}}

 v_2 =\sqrt{\dfrac{4GM}{15R}}

6 0
3 years ago
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